Question

How do you solve $\sin x = \cos \left( {x + 50} \right)$?

Answer 1

Hint: In order to solve the above trigonometric equation, rewrite the right-hand side of the equation with the help of the rule of trigonometry \[\cos x = \sin \left( {\dfrac{\pi }{2} - x} \right)\] by considering $x$ as $\left( {x + 50} \right)$.Take inverse of sine to remove sine from both of the sides . Now combine all the like terms to get the required solution.

Complete step by step solution:
We are given a trigonometric equation $\sin x = \cos \left( {x + 50} \right)$.
$\sin x = \cos \left( {x + 50} \right)$
In order to solve this equation, we will be rewriting the right-hand side of the equation using the rule of trigonometry that \[\cos x = \sin \left( {\dfrac{\pi }{2} - x} \right)\] by considering $x$ in this rule as $\left( {x + 50} \right)$. Out equation now becomes
 \[
  \sin x = \sin \left( {\dfrac{\pi }{2} - \left( {x + 50} \right)} \right) \\
  \sin x = \sin \left( {\dfrac{\pi }{2} - x - 50} \right) \;
 \]
Taking both side inverse of sine, we have
 \[{\sin ^{ - 1}}\left( {\sin x} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{\pi }{2} - x - 50} \right)} \right)\]
Since ${\sin ^{ - 1}}\left( {\sin } \right) = 1$ as they both are inverse of each other and $\dfrac{\pi }{2} = {90^ \circ }$
 \[x = 90 - x - 50\]
Now combining like terms by transposing $x$from the RHS to LHS
 \[
  x + x = 90 - 50 \\
  2x = 40 \;
 \]
Dividing both sides of the equation by the coefficient of $x$i.e. 2, we get
 \[
  \dfrac{{2x}}{2} = \dfrac{{40}}{2} \\
  x = 20 \;
 \]
Therefore, the solution of the given trigonometric equation is $x = 20$
So, the correct answer is “$x = 20$”.

Note: 1.You can also convert the left-hand side of the equation using rule \[\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)\]
2. Verify your answer with the use of a calculator.
3. The equivalent degree value of $\dfrac{\pi }{2}$radian is ${90^ \circ }$
 \[\cos x = \sin \left( {\dfrac{\pi }{2} - x} \right)\]